# Question #11cbb

May 8, 2016

${\text{C"_3"H}}_{8}$

#### Explanation:

Start by writing a balanced chemical equation that describes the combustion of a hydrocarbon ${\text{C"_x"H}}_{y}$

${\text{C"_ x"H"_ (y(g)) + (x + y/4)"O"_ (2(g)) -> x"CO"_ (2(g)) + y/2"H"_ 2"O}}_{\left(g\right)}$

Your job here will be to determine the value of $x$ and the value of $y$, which of course means finding the identity of the unknown hydrocarbon.

The idea here is that when two gases are kept under the same conditions for pressure and temperature, the ratio that exists between their moles is equivalent to the ratio that exists between their volumes.

You can prove this by using the ideal gas law equation, which looks like this

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Now, let's say that the conditions for pressure and temperature are $P$ and $T$. If you take ${n}_{1}$ to be the total number of moles of gas present on the reactants' side, you can say that

$P \cdot {V}_{1} = {n}_{1} \cdot R T \to$ the reactants' side

Likewise, if you take ${n}_{2}$ to be the total number of moles of gas present on the products' side, you can say that

$P \cdot {V}_{2} = {n}_{2} \cdot R T \to$ the products' side

Divide these two equations to get

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{P}}} \cdot {V}_{2}} = \frac{{n}_{1} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}{{n}_{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R T}}}}$

which gets you

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{n}_{1} / {n}_{2} = {V}_{1} / {V}_{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ the mole ratio is equivalent to the volume ratio

So, the total number of moles present on the reactants' side will be

${n}_{1} = 1 + \left(x + \frac{y}{4}\right)$

This is the case because $1$ mole of ${\text{C"_x"H}}_{y}$ reacts with $\left(x + \frac{y}{4}\right)$ moles of oxygen gas.

The total number of moles present on the products' side will be

${n}_{2} = x + \frac{y}{2}$

Here, $x$ moles of carbon dioxide and $\frac{y}{2}$ moles of water are produced by the reaction.

This means that you have

$\frac{1 + x + \frac{y}{4}}{x + \frac{y}{2}} = \left(600 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mL"))))/(700color(red)(cancel(color(black)("mL}}}}\right)$

which is equivalent to

$\frac{1 + x + \frac{y}{4}}{x + \frac{y}{2}} = \frac{6}{7}$

At this point, all you have to do is solve this for $x$ and for $y$.

$1 + x + \frac{y}{4} = 6 \implies x = 5 - \frac{y}{4}$

You will have

$5 - \frac{y}{4} + \frac{y}{2} = 7$

$\frac{y}{4} = 2 \implies y = 8$

The value of $x$ will come out to be

$x = 5 - \frac{8}{4} = 3$

Therefore, the molecular formula of this unknown hydrocarbon is

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\text{C"_3"H}}_{8} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ propane